[Math] Does $\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$ carry an almost complex structure

at.algebraic-topologydg.differential-geometryexceptional-groupssymmetric-spaces

Recall the list of irreducible simply connected inner symmetric spaces of compact type in dimension $4k+2$:

  1. Hermitian symmetric spaces (one can write them down explicitly);

  2. Grassmannians of oriented real $p$-planes in $\mathbb{R}^{p+q}$, with $pq=4k+2$;

  3. The 70-dimensional exceptional symmetric space
    $\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$.

Hermitian symmetric spaces are of course complex manifolds, and real Grassmannians carrying almost complex structures were classified by P. Sankaran (PAMS, 1991) and Z. Tang (PAMS, 1994). Thus my question:

Does
$\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$
carry an almost complex structure?

Note that in dimension $4k$ it is known that an irreducible simply connected inner symmetric space of compact type carries an almost complex structure iff it is Hermitian symmetric.

Edit concerning the dimensions multiple of 4: It is on purpose that I did not mention our paper http://fr.arxiv.org/abs/1003.5172 when I first asked the question. Indeed, the method used in that paper to treat the $4k$-dimensional situation (an index-theoretical obstruction) seems uneffective in the $4k+2$-dimensional cases so I expected other methods to tackle this problem.

Let me nevertheless give a short account on how things work in dimension $4k$. The idea is to use the following criterion:

If a compact Riemannian manifold
$(M,g)$ carries a self-dual vector
bundle $E$ such that the index of the
twisted Dirac operator on $\Sigma M \otimes E\otimes
> TM^{\mathbb{C}}$ is odd, then $M$ has
no almost complex structure (in fact $TM$ is not even stably isomorphic to a complex bundle).

Note that the complex spin bundle $\Sigma M$ and $E$ can be locally defined, only their tensor product has to be globally defined. On then checks that this criterion applies to all irreducible inner symmetric spaces of compact type which are neither Hermitian symmetric nor spheres.

Returning to the dimensions $4k+2$, it is easy to check that there exists a priori no bundle satisfying the above criterion. Nevertheless, one can show real Grassmannians and $\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$ do carry
such a bundle $E$ which satisfies all conditions in the criterion except that it is not self-dual. I don't know if this can lead anywhere…

Best Answer

I'm sorry if this is obvious to everyone, but I thought that it was worth mentioning:

I don't know the answer to the question asked, but the answer to the easier question, "Does the space $\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$ carry an $\mathrm{E}_7$-invariant almost complex structure?", is 'no'.

To see this, consider the orthogonal complement of the subalgebra $\mathfrak{su}(8)$ in $\mathfrak{e}_7$, say $V = \mathfrak{su}(8)^\perp$. By inspecting the root diagram of $\mathfrak{e}_7$ (as described by Cartan's representation, not the usual one you see in most books), one sees that this real vector space of dimension $70$ has the property that its complexification is the irreducible complex representation $\Lambda^4(\mathbb{C}^8)$ of $\mathrm{SU}(8)$. Since the complexification of $V$ is irreducible under $\mathrm{SU}(8)$, the action of $\mathrm{SU}(8)$ does not preserve any complex structure on $V$.

Thus, the space $\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$ does not have an $\mathrm{E}_7$-invariant almost complex structure.

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